Question

Math

Posted 4 months ago

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`Water drips out of a hole at the vertex of an upside down cone at a rate of 3 cm^3 per minute. The cone’s height and radius are 2 cm and 1 cm, respectively. At what rate does the height of the water change when the water level is half a centimeter below the top of the cone? The volume of a cone is V = (π/3)*r^2*h, where r is the radius and h is the height of the cone.`

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Answer from Sia

Posted 4 months ago

Solution by Steps

step 1

Given the volume of a cone $V = \frac{\pi}{3} r^2 h$, differentiate both sides with respect to time $t$ to find the related rates

step 2

Applying the chain rule, we get $\frac{dV}{dt} = \frac{\pi}{3} (2rh \frac{dr}{dt} + r^2 \frac{dh}{dt})$

step 3

Given $\frac{dV}{dt} = -3$ cm³/min (since water is dripping out), $h = 1.5$ cm, and $r = \frac{h}{2}$, we substitute these values into the differentiated equation

step 4

Substituting $r = \frac{h}{2}$ into the equation, we get $\frac{dV}{dt} = \frac{\pi}{3} (2 \cdot \frac{h}{2} \cdot h \cdot \frac{dr}{dt} + (\frac{h}{2})^2 \cdot \frac{dh}{dt})$

step 5

Since $r$ is a function of $h$, $\frac{dr}{dt}$ can be expressed in terms of $\frac{dh}{dt}$. Differentiating $r = \frac{h}{2}$ with respect to $t$, we get $\frac{dr}{dt} = \frac{1}{2} \frac{dh}{dt}$

step 6

Substituting $\frac{dr}{dt}$ into the equation from step 4, we have $-3 = \frac{\pi}{3} (2 \cdot \frac{h}{2} \cdot h \cdot \frac{1}{2} \frac{dh}{dt} + (\frac{h}{2})^2 \cdot \frac{dh}{dt})$

step 7

Simplify the equation to solve for $\frac{dh}{dt}$ when $h = 1.5$ cm

step 8

After simplification, we get $-3 = \frac{\pi}{3} (\frac{3}{4} h^2 \frac{dh}{dt})$

step 9

Substitute $h = 1.5$ cm into the equation to find $\frac{dh}{dt}$

step 10

Solving for $\frac{dh}{dt}$, we get $\frac{dh}{dt} = \frac{-3}{\frac{\pi}{3} \cdot \frac{3}{4} \cdot 1.5^2}$

step 11

Calculate the value of $\frac{dh}{dt}$ to find the rate at which the height of the water changes

step 12

The final value of $\frac{dh}{dt}$ is $\frac{dh}{dt} = \frac{-3}{\frac{\pi}{3} \cdot \frac{3}{4} \cdot 1.5^2} = \frac{-3}{\frac{\pi}{4} \cdot 2.25} = \frac{-3}{\frac{9\pi}{4}} = \frac{-3 \cdot 4}{9\pi} = \frac{-12}{9\pi} = \frac{-4}{3\pi}$ cm/min

Answer

$\frac{dh}{dt} = \frac{-4}{3\pi}$ cm/min

Key Concept

Related Rates in Conical Volume Change

Explanation

The rate of change of the height of the water in the cone is found by differentiating the volume formula with respect to time and substituting the given values to solve for $\frac{dh}{dt}$.

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